What is the integral of (3x)/((6-3x)^{1/2)} ?

The integral of (3x)/((6-3x)^{1/2)} is 2/(3^{1/2)}(-x-4)(-x+2)^{1/2}+C

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integral of (3x)/((6-3x)^{1/2)}

integral

\frac{3 x}{( 6 - 3 x ) ^{\frac{1}{2}}}

\frac{2}{3 ^{\frac{1}{2}}} (- x - 4) (- x + 2)^{\frac{1}{2}} + C

Solution steps

\int \frac{3 x}{( 6 - 3 x ) ^{\frac{1}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{( 6 - 3 x ) ^{\frac{1}{2}}} d x

Apply u-substitution

: \int \frac{u - 6}{9 u ^{\frac{1}{2}}} d u

= 3 \cdot \int \frac{u - 6}{9 u ^{\frac{1}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{9} \cdot \int \frac{u - 6}{u ^{\frac{1}{2}}} d u

Expand

\frac{u - 6}{u ^{\frac{1}{2}}} : u^{\frac{1}{2}} - \frac{6}{u ^{\frac{1}{2}}}

= 3 \cdot \frac{1}{9} \cdot \int u^{\frac{1}{2}} - \frac{6}{u ^{\frac{1}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{9} (\int u^{\frac{1}{2}} d u - \int \frac{6}{u ^{\frac{1}{2}}} d u)

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{6}{u ^{\frac{1}{2}}} d u = 12 u^{\frac{1}{2}}

= 3 \cdot \frac{1}{9} (\frac{2}{3} u^{\frac{3}{2}} - 12 u^{\frac{1}{2}})

Substitute back

u = 6 - 3 x

= 3 \cdot \frac{1}{9} (\frac{2}{3} (6 - 3 x)^{\frac{3}{2}} - 12 (6 - 3 x)^{\frac{1}{2}})

Simplify

3 \cdot \frac{1}{9} (\frac{2}{3} (6 - 3 x)^{\frac{3}{2}} - 12 (6 - 3 x)^{\frac{1}{2}}) : \frac{2}{3 ^{\frac{1}{2}}} (- x - 4) (- x + 2)^{\frac{1}{2}}

= \frac{2}{3 ^{\frac{1}{2}}} (- x - 4) (- x + 2)^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{3 ^{\frac{1}{2}}} (- x - 4) (- x + 2)^{\frac{1}{2}} + C

\frac{2 x ^{2}}{5}

(a + b)^{- 2}

\frac{x}{1 - x}

\frac{- x}{( 1 + x )}

(1 - t^{2})^{3}

What is the integral of (3x)/((6-3x)^{1/2)} ?
The integral of (3x)/((6-3x)^{1/2)} is 2/(3^{1/2)}(-x-4)(-x+2)^{1/2}+C